13(z+3)-5(z-3)=4(z-2)+3(z-2)

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Solution for 13(z+3)-5(z-3)=4(z-2)+3(z-2) equation:



13(z+3)-5(z-3)=4(z-2)+3(z-2)
We move all terms to the left:
13(z+3)-5(z-3)-(4(z-2)+3(z-2))=0
We multiply parentheses
13z-5z-(4(z-2)+3(z-2))+39+15=0
We calculate terms in parentheses: -(4(z-2)+3(z-2)), so:
4(z-2)+3(z-2)
We multiply parentheses
4z+3z-8-6
We add all the numbers together, and all the variables
7z-14
Back to the equation:
-(7z-14)
We add all the numbers together, and all the variables
8z-(7z-14)+54=0
We get rid of parentheses
8z-7z+14+54=0
We add all the numbers together, and all the variables
z+68=0
We move all terms containing z to the left, all other terms to the right
z=-68

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